A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å . Taking hc/e = 1.237 × 10 - 6 e V m and the ground state energy of hydrogen atom as - 13.6 e V , the number of lines present in the emission spectrum is

h c λ = 12370 970 - 13.6 + 12.7 = - 13.6 n 2 n 2 = 16 n = 4 Number of lines = n C 2 = 6

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PhysicsModern PhysicsAtomic structureMedium2 minPYQ_2016

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A hydrogen atom in its ground state is irradiated by light of wavelength $970 Å .$ Taking hc/e = $1.237 \times 10^{- 6} e V$ m and the ground state energy of hydrogen atom as $- 13.6 e V$ , the number of lines present in the emission spectrum is

Answer:

Solution:

$\frac{h c}{λ} = \frac{12370}{970}$
$- 13.6 + 12.7 = - \frac{13.6}{n^{2}}$
$n^{2} = 16$
$n = 4$
Number of lines $=^{n} C_{2} = 6$

Stream:JEE_ADVSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2016

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